dstefan
07-02-2008, 07:06 PM
Chapter 6, Problem 10
Show that the value of a plain vanilla European call option satisfies the Black-Scholes PDE. In other words, show that
\frac{\partial C}{\partial t} + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 C}{\partial S^2} + (r-q) S \frac{\partial C}{\partial S} ~-~ r C ~=~ 0,
where C = C(S,t) is given by the Black-Scholes formula.
Solution:
We use the version of the Black-Scholes PDE involving the Greeks, i.e.,
\Theta ~+~ \frac{1}{2} \sigma^2 S^2 \Gamma ~+~ (r-q) S \Delta ~-~ r C ~=~ 0
and substitute
\Delta = e^{-q(T-t)} N(d_1)
\Gamma = \frac{e^{-q(T-t)}}{S \sigma \sqrt{2 \pi (T-t)}} e^{-\frac{d_1^2}{2}}
\Theta = q S e^{-q(T-t)} N(d_1) ~-~ r K e^{-r(T-t)} N(d_2) ~-~ \frac{\sigma S e^{-q(T-t)}}{2 \sqrt{2 \pi (T-t)}} e^{-\frac{d_1^2}{2}}
Then,
\Theta ~+~ \frac{1}{2} \sigma^2 S^2 \Gamma ~+~ (r-q) S \Delta ~-~ r V
= q S e^{-q(T-t)} N(d_1) ~-~ r K e^{-r(T-t)} N(d_2) ~-~ \frac{\sigma S e^{-q(T-t)}}{2 \sqrt{2 \pi (T-t)}} e^{-\frac{d_1^2}{2}} +~ \frac{\sigma^2 S^2}{2} \frac{e^{-q(T-t)}}{S \sigma \sqrt{2 \pi (T-t)}} e^{-\frac{d_1^2}{2}} +~ (r-q) S e^{-q(T-t)} N(d_1) ~-~ r \left( S e^{-q(T-t)} N( d_1) - K e^{-r(T-t)} N(d_2) \right) = 0
where for C we substituted the value given by the Black-Scholes formula, i.e.,
C ~=~ S e^{-q(T-t)} N( d_1) - K e^{-r(T-t)} N(d_2)
Show that the value of a plain vanilla European call option satisfies the Black-Scholes PDE. In other words, show that
\frac{\partial C}{\partial t} + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 C}{\partial S^2} + (r-q) S \frac{\partial C}{\partial S} ~-~ r C ~=~ 0,
where C = C(S,t) is given by the Black-Scholes formula.
Solution:
We use the version of the Black-Scholes PDE involving the Greeks, i.e.,
\Theta ~+~ \frac{1}{2} \sigma^2 S^2 \Gamma ~+~ (r-q) S \Delta ~-~ r C ~=~ 0
and substitute
\Delta = e^{-q(T-t)} N(d_1)
\Gamma = \frac{e^{-q(T-t)}}{S \sigma \sqrt{2 \pi (T-t)}} e^{-\frac{d_1^2}{2}}
\Theta = q S e^{-q(T-t)} N(d_1) ~-~ r K e^{-r(T-t)} N(d_2) ~-~ \frac{\sigma S e^{-q(T-t)}}{2 \sqrt{2 \pi (T-t)}} e^{-\frac{d_1^2}{2}}
Then,
\Theta ~+~ \frac{1}{2} \sigma^2 S^2 \Gamma ~+~ (r-q) S \Delta ~-~ r V
= q S e^{-q(T-t)} N(d_1) ~-~ r K e^{-r(T-t)} N(d_2) ~-~ \frac{\sigma S e^{-q(T-t)}}{2 \sqrt{2 \pi (T-t)}} e^{-\frac{d_1^2}{2}} +~ \frac{\sigma^2 S^2}{2} \frac{e^{-q(T-t)}}{S \sigma \sqrt{2 \pi (T-t)}} e^{-\frac{d_1^2}{2}} +~ (r-q) S e^{-q(T-t)} N(d_1) ~-~ r \left( S e^{-q(T-t)} N( d_1) - K e^{-r(T-t)} N(d_2) \right) = 0
where for C we substituted the value given by the Black-Scholes formula, i.e.,
C ~=~ S e^{-q(T-t)} N( d_1) - K e^{-r(T-t)} N(d_2)